Fix $k$ to be a local field.
Given a affine algebraic variety say irreducible in $k^n$ (e.g. $X = Z(P_1,\ldots P_r)$) and $G = SL_n$ viewed as the usual algebraic group.
The symmetry group of $X$ is defined to be the stabilizer of $X$ under the action of $G$ on the $P_i$'s. For example for a quadric $Q = 0$ the symmetry group is the orthogonal group of $Q$ which is known to be semisimple. My question is to know whether they are other examples of algebraic affine variety which has a semisimple group of symmetry?

3 Answers
3

I am consolidating my comments above into an answer, because the comments were getting too long (and also I want to correct some mistakes in my comments). Let $W$ be a finite dimensional $k$-vector space of dimension $r$. Let $V$ be the finite-dimensional $k$-vector space $\text{Hom}_k(W,W)$. Let $G$ be $\text{SL}(W)$, the group of linear automorphisms $g$ of $W$ with determinant $1$. Let $\rho:G \to \text{SL}(V)$ be the linear action $g\cdot M = gM$. The $G$-orbit $O$ of $\text{Id}_W \in V$ is an isomorphic copy of $G$, just the image of the usual embedding of $\text{SL}(W)$ in $\text{Hom}_k(W,W)$. Let $L:V\to V$ be any linear transformation that maps the orbit $O$ to itself.

Then, since $L$ is linear, it also preserves the open subset that is the "affine cone" over $O$, i.e., the
image of the open immersion étale morphism $$ \mathbb{G}_m \times G \to V, \ (t,g) \mapsto t\cdot g\cdot \text{Id}_W. $$ Of course this image is precisely the open subset $\text{GL}(W)$ inside $V$. Since $L$ preserves this open subset, it also preserves the closed complement $C$, i.e., the set of non-invertible linear transformations from $V$ to itself.

Of course the closed subset $C$ has a singular stratification, which is the same as the stratification of $\text{Hom}_k(W,W)$ by rank. The linear transformation $L$ preserves every stratum in this stratification. The smallest stratum is $\{ 0 \}$. The next smallest stratum is the locus of rank $1$ matrices $w\otimes \chi$, for $w\in W$ and $\chi\in W^\vee$. Every linear transformation that preserves this locus is of the form $L(M) = AMB$ for (not necessarily unique) matrices $A,B\in \text{GL}(W)$.

Of course $L(\text{Id}_W)$ is in the orbit $O$ if and only if $\text{det}(AB)$ equals $1$. Up to scaling $A$ and $B$ by inverse factors (so $\text{det}(AB)$ is still $1$), $L$ is of the form $L(M)=AMB$ for $A,B\in \text{SL}(W)$. Thus the stabilizer of the orbit $O$ is precisely the image of the homomorphism $$\lambda: \text{SL}(W)\times \text{SL}(W)^{\text{opp}} \to \text{SL}(V), \ \ \lambda(A,B)(M) = AMB. $$ Thus, via the obvious embedding of $SL(W)$ in $V=\text{Hom}_k(W,W)$, $SL(W)$ has semisimple stabilizer.

One way to construct these is to take a semisimple group $G$, a faithful representation $G \to SL_n$, and to just look at the orbits of that action. Often these orbits will have the requisite symmetry group.

A generalization of your two examples is provided by taking affine cones on flag varieties. Flag varieties of semisimple algebraic groups are always projective, so they have cones which are affine varieties, and the connected component of the identity of their symmetry group is always the original algebraic group. Your examples can both be viewed in this way.

Cones yes, but I am not so sure about smooth affine varieties. I am not even sure about smooth affine quadrics in dimension $>1$ (can Youssef give a reference?): even the affine plane has infinite-dimensional group of automorphisms.
–
Serge LvovskiJan 12 '13 at 6:49

Did the question ask about smooth affine varieties? If not, it's clear that $x_1^2+x_2^2+\dots + x_n^2-1$ is smooth with symmetry group $O(n)$. And from the question it seems like we're restricting to linear symmetries.
–
Will SawinJan 12 '13 at 15:45

@Serge: Pleny of Will's orbits will be smooth; in fact, many will just be isomorphic to $G$ itself. For instance, let the affine space $V=k^n$ be the finite-dimensional vector space $\text{Hom}_k(W,W)$, for a finite-dimensional vector space $W$. Let $G$ be $\text{SL}(W)$ acting on $V$ by usual left-multiplication. This gives a representation $\text{SL}(W) \to \text{SL}(V)$. The orbit of $\text{Id}_W \in V$ is a copy of $\text{SL}(V)$, the fiber of $1$ under the determinant $\text{det}:W\to k$. Every element of $\text{SL}(V)$ that preserves this orbit preserves the closure. contd.
–
Jason StarrJan 12 '13 at 16:01

cont. Hence, it also preserves every strata in the "singular stratification" (aka rank stratification) of this orbit. In particular, it preserves the smallest strata (other than $\{ 0 \}$), which is the strata of rank $1$ elements in $V$. It is straightforward to see that such an element of $SL(V)$ must be $M\mapsto AMB$ for $A,B\in \text{SL}(W)$ with $\text{det}(AB)=1$. So the stabilizer is precisely $\text{SL}(W)\times \text{SL}(W)$, which is semisimple.
–
Jason StarrJan 12 '13 at 16:04

Typo corrections. First comment: determinant is $\text{det}:V\to k$, not $\text{det}:W\to k$. Second comment: singular stratification of the orbit closure, not of the orbit (which is smooth).
–
Jason StarrJan 12 '13 at 16:07

Dear Jason, thank you for your brilliant answer my first motivation was to obtain an affine variety given by explicit equations which are left invariant by the action of an isotropic semisimple algebraic group $G$. My motivation is to apply strong approximation to the universal cover of $G$ (this is the reason I need semisimplicty because such isogeny that not exists in general).

This problem is to find a kind of "Inverse invariant theory" (in analogy with Inverse Galois theory), that is, given a algebraic group $G$ to find an algebraic variety which has symmetry group equal to $G$, the problem looks extremely difficult but my question was to know if there is such example known, the only examples I know comes for the theory of prehomgeneous spaces (see arxiv paper, Yukie, Prehomogenous spaces and ergodic theory).